# How do you integrate x^3/sqrt(144-x^2)?

Aug 23, 2015

Use substitution with $u = 144 - {x}^{2}$.

#### Explanation:

$\int {x}^{3} / \sqrt{144 - {x}^{2}} \mathrm{dx}$

Let $u = 144 - {x}^{2}$, so that $\mathrm{du} = - 2 x \mathrm{dx}$ and ${x}^{2} = 144 - u$

$\int {x}^{3} / \sqrt{144 - {x}^{2}} \mathrm{dx} = \int {x}^{2} / \sqrt{144 - {x}^{2}} \text{ } x \mathrm{dx}$

$= - \frac{1}{2} \int {x}^{2} / \sqrt{144 - {x}^{2}} \text{ } \left(- 2 x\right) \mathrm{dx}$

$= - \frac{1}{2} \int \frac{144 - u}{\sqrt{u}} \text{ } \mathrm{du}$

$= - \frac{1}{2} \int \left(\frac{144}{\sqrt{u}} - \sqrt{u}\right) \text{ } \mathrm{du}$

$= - \frac{1}{2} \int \left(144 {u}^{- \frac{1}{2}} - {u}^{\frac{1}{2}}\right) \text{ } \mathrm{du}$

$= - \frac{1}{2} \left[288 {u}^{\frac{1}{2}} - \frac{2}{3} {u}^{\frac{3}{2}}\right] + C$

$= - 144 {\left(144 - {x}^{2}\right)}^{\frac{1}{2}} + \frac{1}{3} {\left(144 - {x}^{2}\right)}^{\frac{3}{2}} + C$

$= - 144 \sqrt{144 - {x}^{2}} + \frac{1}{3} {\left(\sqrt{144 - {x}^{2}}\right)}^{3} + C$